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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 2288 | Accepted: 950 |
Description
Emma and Eric are moving to their new house they bought after returning from their honeymoon. Fortunately, they have a few friends helping them relocate. To move the furniture, they only have two compact cars, which complicates everything a bit. Since the furniture does not fit into the cars, Eric wants to put them on top of the cars. However, both cars only support a certain weight on their roof, so they will have to do several trips to transport everything. The schedule for the move is planed like this:
Note, that the group is always staying together so that they can have more fun and nobody feels lonely. Since the distance between the houses is quite large, Eric wants to make as few trips as possible.
Given the weights wi of each individual piece of furniture and the capacities C1 and C2 of the two cars, how many trips to the new house does the party have to make to move all the furniture? If a car has capacity C, the sum of the weights of all the furniture it loads for one trip can be at most C.
Input
The first line contains the number of scenarios. Each scenario consists of one line containing three numbers n, C1 and C2. C1 and C2 are the capacities of the cars (1 ≤ Ci ≤ 100) and n is the number of pieces of furniture (1 ≤ n ≤ 10). The following line will contain n integers w1, …, wn, the weights of the furniture (1 ≤ wi ≤ 100). It is guaranteed that each piece of furniture can be loaded by at least one of the two cars.
Output
The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Then print a single line with the number of trips to the new house they have to make to move all the furniture. Terminate each scenario with a blank line.
Sample Input
2 6 12 13 3 9 13 3 10 11 7 1 100 1 2 33 50 50 67 98
Sample Output
Scenario #1: 2 Scenario #2: 3
题意：n个物品，两个车，车有自己的容量，物品有自己的体积。求最少的运送次数把物品所有运走，注意，两车一同发车。
解析：状态压缩DP，列举两车可以装下物品的全部方案。然后二进制找两车没有运同样物品的方案，然后用这些方案进行推算就出来了。
注意了，状态压缩就是用二进制的每一位来替代每一个物品，所以n个物品的话最大就是n位
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define Max (1<<15)
using namespace std;
int main (void)
{
int t,n,c1,c2,i,j,k,l1,l2,L,cas=1;
int s[11],dp[Max],s1[Max],s2[Max],dis[Max];
scanf("%d",&t);
while(t--&&scanf("%d%d%d",&n,&c1,&c2))
{
for(i=0;i<n;i++)
scanf("%d",&s[i]);
l1=l2=0;
for(i=0;i<(1<<n);i++) //演算全部可能方案
{
k=0;
for(j=0;j<n;j++) //拿这个方法计算出方案中全部物品的值的总和
if(i&(1<<j))
k+=s[j];
if(k<=c1)s1[l1++]=i; //总和小于c1车的容量表示这个方法能够用c1车来实现
if(k<=c2)s2[l2++]=i; //同上
}
L=0;
for(i=0;i<l1;i++) //一一相应比較
for(j=0;j<l2;j++)
if((s1[i]&s2[j])==0) //与运算用来推断两个二进制数是否有某些位同样，即推断是否某些物品被两车都装了
dis[L++]=(s1[i]|s2[j]);
memset(dp,-1,sizeof(dp));
dp[0]=0; //基础方案初始化
for(i=0;i<(1<<n);i++) //遍历全部状态
if(dp[i]>-1) //预算的前提是这个基础点有值
{
for(j=0;j<L;j++) //遍历全部方案
{
if((i&dis[j])==0&&(dp[i|dis[j]]==-1||dp[i|dis[j]]>dp[i]+1))//这里i&dis[j]==0是用来确定当前状态与当前方案没有冲突。冲突是指的当前状态已经用过某物品而这个方法正好要使用这个物品
dp[i|dis[j]]=dp[i]+1;
}
}
printf("Scenario #%d:\n%d\n\n",cas++,dp[(1<<n)-1]);//输出全部物品都被运送完了的状态中记录的值
}
return 0;
}
我认为这个题非常适用于状压DP的学习。我还刚接触，所以不会没有节操地大发厥词来写算法总结，嘎嘎嘎，前两天说。
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原文地址：http://www.cnblogs.com/lcchuguo/p/4877446.html
